A 19.0-L tank of carbon dioxide gas (CO2) is at a pressure of 9.20 * 10 ^5 Pa an
ID: 1293493 • Letter: A
Question
A 19.0-L tank of carbon dioxide gas (CO2) is at a pressure of 9.20 * 10 ^5 Pa and temperature of 23.0 A degree C. (a) Calculate the temperature of the gas in Kelvin. K (b) Use the ideal gas law to calculate the number of moles of gas in the tank. mol (c) Use the periodic table to compute the molecular weight of carbon dioxide, expressing it in grams per mole. g/mol (d) Obtain the number of grams of carbon dioxide in the tank. g (e) A fire breaks out, raising the ambient temperature by 224.0 K while 82.0 g of gas leak out of the tank. Calculate the new temperature and the number of moles of gas remaining in the tank. temperature K number of moles mol (f) Using the ideal gas law, find a symbolic expression for the final pressure, neglecting the change in volume of the tank. (Use the following as necessary: ni, the initial number of moles; nf, the final number of moles; Ti, the initial temperature; Tf, the final temperature; and Pi, the initial pressure.) Pf= (g) Calculate the final pressure in the tank as a result of the fire and leakage. PaExplanation / Answer
(a) Temperature in Kelvin = Temperature in deg C + 273.15
= 23 + 273.15
= 296.15 K
(b) Using pV = nRuT
9.2*105 * 19*10-3 = n*8.314*296.15
So, no. of moles, n = 7.102
(c) Atmoic weight of carbon = 12 and oxygen = 16
So, atomic weight of CO2 = 12 + 2*16 = 44 g/mol
(d) Mass m = 44*7.102 = 312.48 g
(e) Gas remaining in the tank = 312.48 - 82 = 230.48 g
Moles of gas remaining = 230.48/44 = 5.24 moles
New temperature = 224 + 296.15 = 520.15 K
(f) Using PiVi/(niTi) = PfVf/(nfTf) we get
Putting Vi = Vf
we get,
Pf = (nf/ni)(Tf/Ti)Pi
(g) By putting values in above expression
9.2*105 * 19*10-3 /(7.102*296.15) = Pa * 19*10-3/(5.24*520.15)
This gives final presssure Pa = 11.92*105Pa
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